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%I #26 Oct 14 2024 11:22:00
%S 0,3,6,8,14,20,24,32,42,48,56,72,80,91,110,120,130,153,168,180,208,
%T 224,238,264,288,304,330,360,378,405,440,460,480,527,550,576,624,648,
%U 672,720,754,780,832,868,896,943,990,1020,1062,1120,1152,1196,1258,1292,1326,1400
%N Median absolute deviation of {2*k^2 | k=1..n}.
%C The factor 2 in 2*k^2 in the definition is to ensure that median and median absolute deviation are always integers.
%C Conjecture: a(n) ~ (sqrt(3)/4) * n^2.
%C The conjecture is true. See links. - _Sela Fried_, Jul 17 2024.
%H Sela Fried, <a href="/A345318/a345318.pdf">Median absolute deviation</a>, 2024.
%H Sela Fried, <a href="https://arxiv.org/abs/2410.07237">Proofs of some Conjectures from the OEIS</a>, arXiv:2410.07237 [math.NT], 2024. See p. 3.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Median_absolute_deviation">Median absolute deviation</a>
%e For n = 5 the sample is {2*k^2 | k=1..5} = {2, 8, 18, 32, 50}, its median is 18, the absolute deviations from the median are {16, 10, 0, 14, 32}, the median of the absolute deviations is 14, so a(5) = 14.
%e For n = 6 the sample is {2*k^2 | k=1..6} = {2, 8, 18, 32, 50, 72}, its median is (18+32)/2 = 25, the absolute deviations from the median are {23, 17, 7, 7, 25, 47}, the median of the absolute deviations is (17+23)/2 = 20, so a(6) = 20.
%t Table[MedianDeviation[Table[2 k^2, {k, n}]], {n, 56}]
%Y Cf. A001105 (doubled squares), A000982 (their medians).
%K nonn,easy
%O 1,2
%A _Vladimir Reshetnikov_, Jun 13 2021